Tag: Cross Products

Prove or disprove a given vector formula

by RoRi

May 9, 2016

Prove or disprove:

$A \times \big( A \times (A \times B) \big) \cdot C = -\lVert A \rVert^2 A \cdot B \times C.$

Proof. We can compute using the identities we have derived in the previous exercises of this section,

$\begin{align*} A \times \big( A \times (A \times B) \big) &= A \times \big( (B \cdot A) A - (A \cdot A) B \big) \cdot C \\ &= \big( (A \times A)(B \cdot A) - \lVert A \rVert^2 (A \times B) big) \cdot C \\ &= -\lVert A \rVert^2 ((A \times B) \cdot C) \\ &= -\lVert A \rVert^2 ((A \cdot B) \times C). \qquad \blacksquare \end{align*}$

Prove an identity relating scalar triple products of vectors A,B,C

by RoRi

May 9, 2016

Prove the following identity holds for vectors $A,B,C \in \mathbb{R}^3$ .

$(A \times B) \cdot (B \times C) \times (C \times A) = (A \cdot B \times C)^2.$

Proof. From a previous exercise (Section 13.14, Exercise #9(d)), we know

$(A \times B) \cdot (C \times D) = (B \cdot D)(A \cdot C) - (B \cdot C)(A \cdot D).$

So, with $B \times C$ in place of $C$ and $C \times A$ in place of $D$ in this formula we have

$\begin{align*} (A \times B) \cdot \big((B \times C) \times (C \times A) \big) &= (B \cdot C \times A)(A \cdot B \times C) - (B \cdot B \times C)(A \cdot C \times A) \\ &= (B \times C \cdot A)(A \cdot B \times C) \\ &= (A \cdot B \times C)^2.\qquad \blacksquare \end{align*}$

Compute the cross product of given vectors in terms of the unit coordinate vectors

by RoRi

May 5, 2016

Let $A,B,C,D \in \mathbb{R}^3$ such that

$A \times C \cdot B = 5, \quad A \times D \cdot B = 3, \quad C + D = \mathbf{i} + 2 \mathbf{j} + \mathbf{k}, \quad C - D = \mathbf{i} - \mathbf{k}.$

In terms of the unit coordinate vectors $\mathbf{i}, \mathbf{j}, \mathbf{k}$ compute the cross product

$(A\times B) \times (C \times D).$

Using part (a) of the previous exercise and equation (13.10) on page 490 of Apostol ( $A \times B \cdot C = A \cdot B \times C$ ) we compute

$\begin{align*} (A \times B) \times (C \times D) &= (A \times B \cdot D)C - (A \times B \cdot C)D \\ &= (A \cdot B \times C) C - (A \cdot B \times C) D \\ &= (A \cdot C \times B) D - (A \cdot B \times B) C \\ &= 5D - 3C. \end{align*}$

Then we use the other given relations

$\begin{align*} && C + D &= \mathbf{i} + 2 \mathbf{j} + \mathbf{k} \\ \implies && 2D &= 2 \mathbf{j} + 2 \mathbf{k} \\ \implies && 5D &= 5\mathbf{j} + 5 \mathbf{k} \\ && C-D &= \mathbf{i} - \mathbf{k} \\ \implies && 2C &= 2 \mathbf{i} + 2 \mathbf{j} \\ \implies && 3C &= 3 \mathbf{i} + 3 \mathbf{j}. \end{align*}$

Which all implies

$5D - 3C = -3 \mathbf{i} + 2 \mathbf{j} + 5 \mathbf{k}.$

Prove some properties of the scalar triple product

by RoRi

May 5, 2016

Use the properties of the cross product and the dot product to prove the following properties of the scalar triple product.

$(A+B) \cdot (A+B) \times C = 0$ .
$A \cdot B \times C = -B \cdot A \times C$ .
$A \cdot B \times C = -A \cdot C \times B$ .
$A \cdot B \times C = -C \cdot B \times A$ .

Proof. We have
$(A+B) \cdot (A+B) \times C = (A+B) \cdot ((A+B) \times C) = 0$

since $D \cdot (D \times E) = 0$ for any vectors $D,E$ (in this case $D = A+B$ and $E = C). \qquad blacksquare$
Proof. Using part (a), we have
$\begin{align*} (A+B) \cdot ((A+B) \times C) = 0 && \implies && (A+B) \cdot ((A \times C) + (B \times C)) &= 0 \\ && \implies && B \cdot (A \times C) + A \cdot (B \times C) &= 0 \\ && \implies && A \cdot (B \times C) &= -B \cdot (A \times C). \qquad \blacksquare \end{align*}$
Proof. We have
$A \cdot B \times C = A \cdot (-C \times B) = A \cdot (-1)(C \times B) = -A \cdot (C \times B). \qquad \blacksquare$
Proof. We have
$\begin{align*} A \cdot B \times C &= -A \cdot C \times B &(\text{part c}) \\ &= C \cdot A \times B &(\text{part b}) \\ &= -C \cdot B \times A &(\text{part c}). \qquad \blacksquare \end{align*}$

Prove that there is exactly one vector satisfying given conditions

by RoRi

May 2, 2016

Prove that there is one and only one vector $B$ satisfying the equations

$A \times B = C, \qquad A \cdot B = 1$

where $A \neq O$ and $C$ is orthogonal to $A$ in $\mathbb{R}^3$ .

Proof. First, we show uniqueness. If $B$ is any vector such that $A \times B = C$ and $A \cdot B = 1$ and $B'$ is another vector such that $A \times B' = C$ and $A \cdot B' = 1$ then we have

$A \times B = A \times B', \qquad A \cdot B = A \cdot B'.$

But, by a previous exercise (Section 13.11, Exercise #8) we know these conditions imply $B = B'$ . Hence, there is at most one such $B$ .

Now, for existence. Let $A = (a_1, a_2, a_3), \ B = (b_1, b_2,b_3), \ C = (c_1, c_2, c_3)$ . Since $A,C$ are orthogonal we know $A \cdot C = 0$ ,

$a_1 c_1 + a_2 c_2 + a_3 c_3 = 0.$

From $A \times B = C$ we then have

$\begin{align*} a_2 b_3 - a_3 b_2 &= c_1 \\ a_3 b_1 - a_1 b_3 &= c_2 \\ a_1 b_2 - a_2 b_1 &= c_3. \end{align*}$

Since $A \neq O$ , we know at least one of the $a_i \neq 0$ . Without loss of generality, assume $a_1 \neq 0$ . From the second and third equations we have

$b_3 = \frac{a_3 b_1 - c_2}{a_1}, \qquad b_2 = \frac{c_3 + a_2 b_1}{a_1}.$

This implies

$\begin{align*} a_2 b_3 - a_3 b_2 = c_1 && \implies && \frac{a_2 a_3 b_1 - a_2 c_2}{a_1} - \frac{a_3 c_3 + a_3 a_2 b_1}{a_1} &= c_1 \\[9pt] && \implies && a_2 a_3 b_1 + a_3 a_2 b_1 - a_3 c_3 + a_3 a_2 b_1 &= a_1 c_1 \\[9pt] && \implies && a_2 a_3 b_1 + a_3 a_2 b_1 &= a_1 c_1 + a_2 c_2 + a_3 c_3 \\[9pt] && \implies && (a_2 a_3 + a_3 a_2)b_1 &= 0 \\ && \implies && a_2 a_3 + a_3 a_2 &= 0. \end{align*}$

Hence, $b_1$ can take any value. The vectors $B$ such that $A \times B = C$ are then of the form

$B = \left( b_1, \frac{c_3 + a_2 b_1}{a_1}, \frac{a_3 b_1 - c_2}{a_1} \right).$

Then,

$\begin{align*} A \cdot B = 1 && \implies && a_1 b_1 + a_2 \left( \frac{c_2 + a_2 b_1}{a_1} \right) + a_3 \left( \frac{a_3 b_1 - c_2}{a_1} \right) &= 1 \\[9pt] && \implies && a_1^2 b_1 + a_2 c_3 + a_2^2 b_1 + a_3^2 b_1 - a_3 c_2 &= 1\\[9pt] && \implies && b_1 &= \left( \frac{1}{a_1^2 + a_2^2 + a_3^2} \right) (a_3 c_2 - a_2 c_3). \end{align*}$

Since $a_1^2 + a_2^2 + a_3^2 \neq 0$ (since at least one of the $a_i \neq 0$ , we have that such a vector $B$ always exists $. \qquad \blacksquare$

Find a vector that is equal to the cross product of two given vectors

by RoRi

May 2, 2016

Consider the vectors

$A = 2 \mathbf{i} - \mathbf{j} + 2 \mathbf{k}, \qquad C = 3 \mathbf{i} + 4 \mathbf{j} - \mathbf{k}.$

Find $B$ satisfying $A \times B = C$ . How many such solutions are there?
Find $B$ such that $A \times B = C$ and $A \cdot B = 1$ . How many such solutions are there?

Let $B = (b_1, b_2, b_3)$ . For $A \times B = C$ we must have
$\begin{align*} && \begin{vmatrix*} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 2 \\ b_1 & b_2 & b_3 \end{vmatrix*} &= 3 \mathbf{i} + 4 \mathbf{j} - \mathbf{k} \\[9pt] \implies && \begin{vmatrix*} -1 & 2 \\ b_2 & b_3 \end{vmatrix*} \mathbf{i} - \begin{vmatrix*} 2 & 2 \\ b_1 & b_3 \end{vmatrix*} \mathbf{j} + \begin{vmatrix*} 2 & -1 \\ b_1 & b_2 \end{vmatrix*} \mathbf{k} &= 3 \mathbf{i} + 4 \mathbf{j} - \mathbf{k} \\[9pt] \implies && (-b_3 - 2b_2) \mathbf{i} + (2b_1 - 2b_3) \mathbf{j} + (2b_2 + b_1) \mathbf{k} &= 3 \mathbf{i} + 4 \mathbf{j} - \mathbf{k}. \end{align*}$

Therefore, we have the three equations,

$\begin{align*} -b_3 - 2b_2 &= 3 \\ 2b_1 - 2b_3 &= 4 \\ 2b_2 + b_1 &= -1. \end{align*}$

From the first equation we have $b_3 = -3-2b_2$ . From the second equation we then have $b_1 = -1 - 2b_2$ . Since any value of $b_2$ then satisfies the third equation we have that $b_2$ is arbitrary. Letting $b_2 = 0$ we then have $b_1 = -1$ and $b_3 = -3$ . Hence, a solution is

$B = (-1,0,-3) = -\mathbf{i} - 3\mathbf{k}.$

There are infinitely many solutions since we can take any value for $b_2$ to obtain another solution.
From part (a) we know that the vectors $B$ such that $A \times B = C$ are of the form
$B = (-1-2b_2, b_2, -3-2b_2)$

for any value of $b_2$ . Then,

$\begin{align*} && A \cdot B &= 1 \\ \implies && 2 \cdot (-1-2b_2) + (-1)(b_2) + 2(-3-2b_2) &= 1 \\ \implies && -2-4b_2 -b_2 - 6 - 4b_2 &=1 \\ \implies && -8-9b_2 &= 1 \\ \implies && b_2 &= -1. \end{align*}$

From part (a) we then have $b_1 = 1$ and $b_3 = -1$ . Hence, $B = \mathbf{i} - \mathbf{j} - \mathbf{k}$ is the only solution.

Prove some facts about orthogonal unit vectors

by RoRi

May 2, 2016

Let $A,B \in \mathbb{R}^3$ be orthogonal unit vectors.

Prove that the vectors $A, \ B, A \times B$ form an orthonormal basis for $\mathbb{R}^3$ .
Prove that the vector $C = (A \times B) \times A$ has unit length.
Show the geometric relation between $A, \ B, A \times B$ and obtain the relations
$(A \times B) \times A = B, \qquad (A \times B)\times B = -A.$
Prove the relations in part (c) algebraically.

Proof. By Theorem 13.13 (page 484 of Apostol) we know that if $A$ and $B$ are independent then so is the set $\{ A, B, A \times B \}$ . We also know by Theorem 13.12 (page 483) that $A \times B$ is orthogonal to both $A$ and $B$ . Since this is a set of three independent vectors in $\mathbb{R}^3$ , it is a basis. Then, if $A,B$ each have length 1 and are orthogonal we have
$\lVert A \times B \rVert = \lVert A \rVert \lVert B \rVert \sin \theta = (1)(1)\left( \sin \frac{\pi}{2} \right) = 1.$

Hence, $A \times B$ has length 1 as well. Thus, $\{ A, B, A \times B \}$ is form an orthonormal basis $. \qquad \blacksquare$
Proof. From part (a) we know that $(A \times B)$ and $A$ each have length 1 and are orthogonal. Thus,
$\lVert C \rVert^2 = \lVert A \times B \rVert^2 \lVert A \rVert^2 \sin \frac{\pi}{2} = (1)(1)(1) = 1. \qquad \blacksquare$
Incomplete.
Proof. Since $A,B,A \times B$ are orthogonal, we know every vector orthogonal to two of them is a scalar multiple of the third. Thus, $(A \times B) \times A$ is a scalar multiple of $B$ . Further, since $A, B, A \times B$ all have length 1
$\lVert (A \times B) \times A \rVert = \lVert cB \rVert \quad \implies \quad c = \pm 1.$

If we adopt a right hand coordinate system then $c = 1$ . So, $(A \times B) \times A = B$ . Similarly, $(A \times B) \times B = -A. \qquad \blacksquare$

Prove that the norm of the cross product is the product of the norms if and only if A and B are orthogonal

by RoRi

May 1, 2016

Given vectors $A,B$ prove that

$\lVert A \times B \rVert =\lVert A \rVert \lVert B \rVert$

if and only if $A$ and $B$ are orthogonal.

Proof.

From Theorem 13.12(f) (page 483 of Apostol) we know

$\lVert A \times B \rVert^2 = \lVert A \rVert^2 + \lVert B \rVert^2 - (A \cdot B)^2.$

But $(A \cdot B)^2 = 0$ if and only if $A$ and $B$ are orthogonal (from the definition of orthogonality). Thus, $\lVert A \times B \rVert = \lVert A \rVert \lVert B \rVert$ if and only if $A$ and $B$ are orthogonal $. \qquad \blacksquare$

Express the cross product of given vectors in terms of the unit coordinate vectors

by RoRi

May 1, 2016

Let

$A = 2 \mathbf{i} + 5 \mathbf{j} + 3 \mathbf{k}, \quad B = 2 \mathbf{i} + 7 \mathbf{j} + 4 \mathbf{k}, \quad C = 3 \mathbf{i} + 3 \mathbf{j} + 6 \mathbf{k}.$

Find $(A - C) \times (B - A)$ in terms of $\mathbf{i}, \mathbf{j}, \mathbf{k}$ .

First, we have

$\begin{align*} A - C &= - \mathbf{i} + 2\mathbf{j} - 3\mathbf{k} \\ B - A &= 2\mathbf{j} + \mathbf{k}. \end{align*}$

So, the cross product is given by

$\begin{align*} (A - C) \times (B - A) &= \begin{vmatrix*} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 2 & -3 \\ 0 & 2 & 1 \end{vmatrix*} \\[9pt] &= \begin{vmatrix*} 2 & -3 \\ 2 & 1 \end{vmatrix*} \mathbf{i} - \begin{vmatrix*} -1 & -3 \\ 0 & 1 \end{vmatrix*} \mathbf{j} + \begin{vmatrix*} -1 & 2 \\ 0 & 2 \end{vmatrix*} \mathbf{k} \\[9pt] &= 8 \mathbf{i} + \mathbf{j} - 2 \mathbf{k} \end{align*}$

Use the cross product to compute the area of triangles with given vertices

by RoRi

May 1, 2016

For each of the following sets of points $A,B,C$ use the cross product to compute the area of the triangle with vertices $A,B,C$ .

$A = (0,2,2), \quad B = (2,0,-1), \quad C = (3,4,0)$ ;
$A = (-2,3,1), \quad B = (1,-3,4), \quad C = (1,2,1)$ ;
$A = (0,0,0), \quad B = (0,1,1), \quad C = (1,0,1)$ .

The area of the triangle with vertices $A,B,C$ is given by
$\begin{align*} \text{Area} &= \frac{1}{2} \lVert (A - C) \times (B-C) \rVert \\ &= \frac{1}{2} \lVert (-3,-2,2) \times (-1,-4,-1) \rVert \\ &= \frac{1}{2} \lVert (10,-5,10) \rVert \\ &= \frac{\sqrt{225}}{2} \\ &= \frac{15}{2}. \end{align*}$
The area of the triangle with vertices $A,B,C$ is given by
$\begin{align*} \text{Area} &= \frac{1}{2} \lVert (A - C) \times (B-C) \rVert \\ &= \frac{1}{2} \lVert (-3,1,0) \times (0,-5,3) \rVert \\ &= \frac{1}{2} \lVert (3,9,15) \rVert \\ &= \frac{\sqrt{333}}{2} \\ &= \frac{3}{2} \sqrt{37}. \end{align*}$
The area of the triangle with vertices $A,B,C$ is given by
$\begin{align*} \text{Area} &= \frac{1}{2} \lVert (A - C) \times (B-C) \rVert \\ &= \frac{1}{2} \lVert (-1,0,-1) \times (-1,1,0) \rVert \\ &= \frac{1}{2} \lVert (1,1,-1) \rVert \\ &= \frac{\sqrt{3}}{2}. \end{align*}$